In mathematics, algebras A, B over a field k inside some field extension
Ω
of k are said to be linearly disjoint over k if the following equivalent conditions are met:
(i) The map
A
⊗
k
B
→
A
B
induced by
(
x
,
y
)
↦
x
y
is injective.
(ii) Any k-basis of A remains linearly independent over B.
(iii) If
u
i
,
v
j
are k-bases for A, B, then the products
u
i
v
j
are linearly independent over k.
Note that, since every subalgebra of
Ω
is a domain, (i) implies
A
⊗
k
B
is a domain (in particular reduced). Conversely if A and B are fields and either A or B is an algebraic extension of k and
A
⊗
k
B
is a domain then it is a field and A and B are linearly disjoint. However there are examples where
A
⊗
k
B
is a domain but A and B are not linearly disjoint: for example, A=B=k(t), the field of rational functions over k.
One also has: A, B are linearly disjoint over k if and only if subfields of
Ω
generated by
A
,
B
, resp. are linearly disjoint over k. (cf. tensor product of fields)
Suppose A, B are linearly disjoint over k. If
A
′
⊂
A
,
B
′
⊂
B
are subalgebras, then
A
′
and
B
′
are linearly disjoint over k. Conversely, if any finitely generated subalgebras of algebras A, B are linearly disjoint, then A, B are linearly disjoint (since the condition involves only finite sets of elements.)
